Trigonometry · The Lumenary

MotivationH

Measuring what you cannot reach

How do you measure the height of a mountain you cannot climb, or the distance to a ship you cannot reach? In the 19th century the Great Trigonometrical Survey of India set out to map the subcontinent using nothing but carefully measured baselines and angles, observed with heavy theodolites from hilltops and towers. Among the "computers" who turned those observations into distances and heights was the Bengali mathematician Radhanath Sikdar. He is widely credited with the calculations, around 1852, showing that a distant Himalayan summit known as Peak XV was the highest mountain on Earth. In 1856 the Survey announced its height as $29{,}002$ feet. We now call it Everest, or Sagarmāthā in Nepali and Chomolungma in Tibetan.

The key fact is simple: in a right triangle, an angle determines the ratios of the sides. Measure one length and one angle, and trigonometry gives you the others. The same functions — sine and cosine — also describe anything that repeats: tides, sound waves, alternating current, the seasons. In this lesson we define them, prove the identities that connect them, and derive the two laws that solve any triangle.

IntuitionH

Why a ratio can belong to an angle

Take any two right triangles that share an acute angle $\theta$ . Their third angles are both $90^\circ - \theta$ , so they are similar (AA; see rigid motions, congruence and similarity). One is a scaled copy of the other, so every ratio of sides is the same in both:

\frac{\text{opposite}}{\text{hypotenuse}} \text{ is the same for every right triangle with angle } \theta.

That ratio depends only on $\theta$ , so we can give it a name and tabulate it once and for all. This is exactly what astronomers in Greece, India and the Islamic world did, centuries before calculators: they made tables, and anyone could then convert angles into lengths.

DefinitionsH

Definitions

Definition— sine, cosine and tangent of an acute angle

In a right triangle with an acute angle $\theta$ , let the hypotenuse be the side opposite the right angle, the opposite side be the side facing $\theta$ , and the adjacent side be the other side touching $\theta$ . Then

\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}, \qquad \cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}, \qquad \tan\theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{\sin\theta}{\cos\theta}.

Angles can be measured in degrees, but mathematics prefers a natural unit that comes from the circle itself.

Definition— radian

An angle of one radian at the centre of a circle cuts off an arc whose length equals the radius. In general, an angle of $\theta$ radians cuts off an arc of length $s = r\theta$ on a circle of radius $r$ . A full turn is an arc of length $2\pi r$ , so

2\pi \text{ radians} = 360^\circ, \qquad \pi \text{ radians} = 180^\circ, \qquad 1 \text{ radian} = \frac{180^\circ}{\pi} \approx 57.3^\circ.

To convert degrees to radians, multiply by $\pi/180$ ; to convert back, multiply by $180/\pi$ . So $90^\circ = \pi/2$ , $60^\circ = \pi/3$ , $45^\circ = \pi/4$ and $30^\circ = \pi/6$ .

The triangle definition only works for angles between $0^\circ$ and $90^\circ$ . To go further, we use a circle.

Definition— sine and cosine of any angle (the unit circle)

The unit circle has centre $O = (0, 0)$ and radius $1$ . Start at $(1, 0)$ and turn through an angle $\theta$ about $O$ — anticlockwise if $\theta > 0$ , clockwise if $\theta \lt 0$ . The point you reach is, by definition,

(\cos\theta, \sin\theta).

Wherever $\cos\theta \ne 0$ , we define $\tan\theta = \sin\theta / \cos\theta$ , the slope of the line from $O$ to that point.

For $0 \lt \theta \lt 90^\circ$ the two definitions agree: drop a perpendicular from the point $(\cos\theta, \sin\theta)$ to the $x$ -axis. You get a right triangle with hypotenuse $1$ , adjacent side $\cos\theta$ and opposite side $\sin\theta$ .

Because turning through $2\pi$ brings you back to where you started, $\sin$ and $\cos$ are periodic with period $2\pi$ (that is, $360^\circ$ ):

\sin(\theta + 2\pi) = \sin\theta, \qquad \cos(\theta + 2\pi) = \cos\theta.

Reflecting in the $x$ -axis turns $\theta$ into $-\theta$ , so $\sin(-\theta) = -\sin\theta$ (sine is odd) and $\cos(-\theta) = \cos\theta$ (cosine is even). The tangent has period $\pi$ and is undefined where $\cos\theta = 0$ , at $\theta = \pi/2 + k\pi$ .

Worked examplesH

Worked examples

Example— Exact values from two special triangles

Cut a square of side $1$ along a diagonal: you get a right triangle with angles $45^\circ, 45^\circ, 90^\circ$ and sides $1, 1, \sqrt2$ . Cut an equilateral triangle of side $2$ in half: you get angles $30^\circ, 60^\circ, 90^\circ$ and sides $1, \sqrt3, 2$ (by Pythagoras, $\sqrt{2^2 - 1^2} = \sqrt3$ ). Reading off the ratios:

$\theta$	$0$	$30^\circ = \pi/6$	$45^\circ = \pi/4$	$60^\circ = \pi/3$	$90^\circ = \pi/2$
$\sin\theta$	$0$	$\tfrac12$	$\tfrac{\sqrt2}{2}$	$\tfrac{\sqrt3}{2}$	$1$
$\cos\theta$	$1$	$\tfrac{\sqrt3}{2}$	$\tfrac{\sqrt2}{2}$	$\tfrac12$	$0$
$\tan\theta$	$0$	$\tfrac{\sqrt3}{3}$	$1$	$\sqrt3$	undefined

Example— Angles in every quadrant

The point at $150^\circ$ on the unit circle is the mirror image, in the $y$ -axis, of the point at $30^\circ$ . So $\sin 150^\circ = \sin 30^\circ = \tfrac12$ and $\cos 150^\circ = -\cos 30^\circ = -\tfrac{\sqrt3}{2}$ . In the same way, $240^\circ = 180^\circ + 60^\circ$ is the point at $60^\circ$ reflected through $O$ , so $\cos 240^\circ = -\tfrac12$ and $\sin 240^\circ = -\tfrac{\sqrt3}{2}$ . And $\sin(-30^\circ) = -\tfrac12$ because sine is odd.

In general $\sin(180^\circ - \theta) = \sin\theta$ : an angle and its supplement have the same sine. This will matter for the law of sines.

Example— The height of a tower you cannot reach

From a point $A$ on level ground, the top of a tower is seen at an angle of elevation of $30^\circ$ . After walking $100$ m straight towards the tower, to $B$ , the angle is $45^\circ$ . How tall is the tower?

Let the height be $h$ and the distance from $B$ to the foot of the tower be $x$ . From $B$ : $\tan 45^\circ = h / x$ , so $h = x$ . From $A$ : $\tan 30^\circ = h / (x + 100)$ , so $x + 100 = \sqrt3\,h$ . Hence $h + 100 = \sqrt3\,h$ , and

h = \frac{100}{\sqrt3 - 1} = 50\left(\sqrt3 + 1\right) \approx 136.6 \text{ m}.

Surveyors measuring a mountain do the same thing on a grander scale, with corrections for the curvature of the Earth and the bending of light in the air.

Example— Radians in action

A bicycle wheel of radius $0.35$ m turns through $3$ radians. A point on the rim travels $s = r\theta = 0.35 \times 3 = 1.05$ m. In degrees the same angle is $3 \times 180^\circ/\pi \approx 171.9^\circ$ , and the formula would need an awkward factor: $s = r\theta\pi/180$ . Radians make arc length, and (in calculus) the derivatives of $\sin$ and $\cos$ , as simple as possible.

ProofH

The identities and the laws

Theorem— Pythagorean identity

For every angle $\theta$ : $\;\sin^2\theta + \cos^2\theta = 1$ .

Proof

The point $(\cos\theta, \sin\theta)$ is on the unit circle, so its distance from $O$ is $1$ . By the distance formula (which is the Pythagorean theorem; see the Pythagorean theorem), $\cos^2\theta + \sin^2\theta = 1^2$ . The distance formula squares the coordinates, so their signs do not matter: the identity holds in every quadrant. $\blacksquare$

∎

Here $\sin^2\theta$ means $(\sin\theta)^2$ . Dividing the identity by $\cos^2\theta$ gives $1 + \tan^2\theta = 1/\cos^2\theta$ wherever $\cos\theta \ne 0$ .

Theorem— Addition formulas

For all angles $\alpha$ and $\beta$ :

\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta, \qquad \cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta.

Proof

Let $R_\theta$ be the rotation of the plane about $O$ through the angle $\theta$ . We use three geometric facts about rotations.

Rotations add. Rotating by $\beta$ and then by $\alpha$ is rotating by $\alpha + \beta$ . In particular, rotations about $O$ commute.
A quarter turn. Rotating the point $(x, y)$ by $90^\circ$ gives $(-y, x)$ . (Rotate the right triangle with vertices $O$ , $(x, 0)$ and $(x, y)$ by a quarter turn.)
Rotations respect scaling and parallelograms. A rotation about $O$ is a rigid motion fixing $O$ , so it sends the point $t\,P$ to $t\,R(P)$ and the parallelogram with vertices $O$ , $P$ , $Q$ , $P + Q$ to a parallelogram. Hence $R(xP + yQ) = x\,R(P) + y\,R(Q)$ .

By the definition of sine and cosine, $R_\alpha(1, 0) = (\cos\alpha, \sin\alpha)$ . By facts 1 and 2,

R_\alpha(0, 1) = R_\alpha R_{90^\circ}(1, 0) = R_{90^\circ} R_\alpha(1, 0) = R_{90^\circ}(\cos\alpha, \sin\alpha) = (-\sin\alpha, \cos\alpha).

Now write $(\cos\beta, \sin\beta) = \cos\beta\,(1, 0) + \sin\beta\,(0, 1)$ and use fact 3:

R_\alpha(\cos\beta, \sin\beta) = \cos\beta\,(\cos\alpha, \sin\alpha) + \sin\beta\,(-\sin\alpha, \cos\alpha) = (\cos\alpha\cos\beta - \sin\alpha\sin\beta,\; \sin\alpha\cos\beta + \cos\alpha\sin\beta).

But $(\cos\beta, \sin\beta) = R_\beta(1, 0)$ , so by fact 1 the left side is $R_{\alpha + \beta}(1, 0) = (\cos(\alpha + \beta), \sin(\alpha + \beta))$ . Comparing coordinates gives both formulas. $\blacksquare$

∎

Theorem— Corollaries

$\sin(\alpha - \beta) = \sin\alpha\cos\beta - \cos\alpha\sin\beta$ and $\cos(\alpha - \beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta$ .
Double angles: $\sin 2\theta = 2\sin\theta\cos\theta$ and $\cos 2\theta = \cos^2\theta - \sin^2\theta = 1 - 2\sin^2\theta = 2\cos^2\theta - 1$ .

Proof

Replace $\beta$ by $-\beta$ and use $\sin(-\beta) = -\sin\beta$ , $\cos(-\beta) = \cos\beta$ .
Put $\alpha = \beta = \theta$ , then use $\sin^2\theta + \cos^2\theta = 1$ . $\blacksquare$

∎

For example, $\sin 75^\circ = \sin(45^\circ + 30^\circ) = \tfrac{\sqrt2}{2}\cdot\tfrac{\sqrt3}{2} + \tfrac{\sqrt2}{2}\cdot\tfrac12 = \tfrac{\sqrt6 + \sqrt2}{4} \approx 0.966$ .

From now on, in a triangle $ABC$ we write $a = BC$ , $b = CA$ , $c = AB$ for the sides, and $A$ , $B$ , $C$ also for the angles at those vertices.

Theorem— Law of cosines

In every triangle, $\;c^2 = a^2 + b^2 - 2ab\cos C$ .

Proof

Place $C$ at the origin and $B$ at $(a, 0)$ on the positive $x$ -axis, with $A$ above the axis. The point $A$ is at distance $b$ from $C$ , in the direction at angle $C$ from the positive $x$ -axis, so by the unit-circle definition $A = (b\cos C, b\sin C)$ . This holds whether $C$ is acute, right or obtuse. By the distance formula,

c^2 = (b\cos C - a)^2 + (b\sin C)^2 = b^2\left(\cos^2 C + \sin^2 C\right) - 2ab\cos C + a^2 = a^2 + b^2 - 2ab\cos C. \qquad \blacksquare

∎

When $C = 90^\circ$ , $\cos C = 0$ and the law of cosines is the Pythagorean theorem. When $C$ is obtuse, $\cos C \lt 0$ and $c^2 > a^2 + b^2$ : the side opposite an obtuse angle is longer than Pythagoras would give.

Theorem— Law of sines

In every triangle, $\;\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$ .

Proof

In the coordinates of the last proof, the height of $A$ above the line $CB$ is $b\sin C$ , so the area of the triangle is

\text{Area} = \tfrac12 \cdot a \cdot b\sin C.

The same argument at the other vertices gives $\text{Area} = \tfrac12 bc\sin A = \tfrac12 ca\sin B$ . So $\tfrac12 ab\sin C = \tfrac12 bc\sin A = \tfrac12 ca\sin B$ . Dividing by $\tfrac12 abc$ gives $\dfrac{\sin C}{c} = \dfrac{\sin A}{a} = \dfrac{\sin B}{b}$ , and taking reciprocals (the sines of the angles of a triangle are positive) gives the law. $\blacksquare$

∎

Note

The common ratio has a meaning: $\dfrac{a}{\sin A} = 2R$ , where $R$ is the radius of the circumcircle. For acute $A$ , draw the diameter $BD$ of the circumcircle. By the inscribed angle theorem $\angle BDC = A$ , and by Thales' theorem the triangle $BCD$ has a right angle at $C$ ; so $a = BC = BD \sin A = 2R\sin A$ . (See Euclidean geometry and proof.)

Note

The ambiguous case returns. If you know $a$ , $b$ and $A$ and use the law of sines to find $B$ , you get $\sin B = b\sin A / a$ . Since $\sin B = \sin(180^\circ - B)$ , there may be two possible angles — exactly the SSA ambiguity from Euclidean geometry. Always check whether $180^\circ - B$ also fits in the triangle.

Formal mathematicsF

Trigonometry in Lean

In Mathlib, Real.sin and Real.cos are not defined by triangles at all: they are defined by power series, through the complex exponential function, and they always take radians. So we begin with a conversion from degrees, which plays the role of a calculator's degree mode:

degLean 4Formally verified

/-- Convert degrees to radians: `x°` is `x · π / 180` radians. Mathlib's `Real.sin` always
expects radians, exactly like a calculator in radian mode. -/
noncomputable def deg (x : ℝ) : ℝ := x * π / 180